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3 years ago

Find the measure of arc CE.

Mathematics

2 answers:

almond37 [142]3 years ago

7 0

∠CAE is is the central angle ⇒ arc CE = ∠CAE
∠СAE = ∠СAD + ∠DAE = 40° + 25° = 65°
So arc CE = 65°

kati45 [8]3 years ago

6 0

Arc CE = 40 + 25 = 65
answer
65

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$\large\underline{\sf{Solution-}}$

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Hence, 

$\bf :\longmapsto\:\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } = \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg]$

Therefore, 

$\purple{\rm :\longmapsto\:\displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty }\rm \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg]$

$\rm \: = \: \rm \dfrac{1}{2}\bigg[1 - 0 \bigg]$

$\rm \: = \: \rm \dfrac{1}{2}$

Hence, 

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]} = \frac{1}{2}}}$

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3 years ago

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Bogdan [553]

Answer:

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Step-by-step explanation:

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z <(x - m) / sd

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replacing:

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